1.

Calculatemolefractionof HCIin asolutioncontaining24.8% of HCIby mass.

Answer»


Solution :Given : %by weightofHCI = 24.8 ,
Molefractionof HCI= `X_(HCI)= ?`
Consider100 gramHCIsolution .
`:.` Weightof a solute(HCI) =24.8 G
Weightof ASOLVENT(water)= 100 -24.8 = 75.2 g
`:. ` Molesof solvent= `n_(H_(2)O) = .(W_(H_(2)O))/(M_(H_(2)O))=(75.2)/(18) =4.178 mol H_(2)O`
Molesof solute`=n_(H_(2)O) =(W_(H_(2)O))/(36.5) =(24.8)/(36.5)=0.6795 mol`
`:. ` Totalmolesin solution`= n_(H_(2)O) + n_(HCI)`
`=4. 178+ 0.6795 =4. 8575`
Molefractionof HCI `X_("HCI") =("molesof HCI")/("totalmolesin solution")`
`= (0.6795)/(4.8575)= 0.1398`


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